We know that the monodromy group of the space of \(y^2 = f(x)\)  with \(f(x) \in \mathcal {H}_{d_1, \dots , d_n}\)  is a subgroup of \(\mathrm {Sp}(2g)\) . We would like to show that this monodromy group is in fact the full group. However, the monodromy group of a pullback is a subgroup of the original monodromy group, so it suffices to show that after an appropriate choice of pullback, we have the full monodromy group \(\mathrm {Sp}(2g)\) .
To do that, we fix a polynomial \(f_{2g}(x) \in \mathcal {H}_{d_1 -1, \dots , d_n}\) . Consider the space of \(y^2 = f(x)\)  with \(f(x) = (x-t)f_{2g}(x)\) , with a single parameter \(t\) . We restrict this space to the open subset of \(t\)  where \(f_{2g}(t) \ne 0\) . We now would like to map this restricted space to the space of \(y^2=f(x)\)  with \(f(x) \in \mathcal {H}_{d_1, \dots , d_n}\) , which we do by multiplying the first component by the \((x-t)\)  factor. As mentioned above, the monodromy group of the space of \(y^2 = f(x)\)  with \(f(x) = (x-t)f_{2g}(x)\)  is a subgroup of the monodromy group that we would like to understand. However, by [1]} (see also [2]} and [3]}), the monodromy group of our one-parameter space is already the full group \(\mathrm {Sp}(2g)\) .
This implies that the family \(y^2 = f(x)\)  with \(f(x) \in \mathcal {H}_{d_1, \dots , d_n}\)  has monodromy group \(\mathrm {Sp}(2g)\) . By Theorem 9.2.6 in [1]}, the large monodromy of this family implies the equidistribution statement.
